1. The document discusses remote sensing applications and techniques such as linear contrast stretching, histogram equalization stretching, calculating slope and radiance from metadata, and calculating land surface temperature from at-sensor brightness temperature. 2. Formulas are provided for calculating land surface emissivity based on NDVI and proportion of vegetation, as well as calculating land surface temperature using parameters like brightness temperature, wavelength, and emissivity. 3. Worked examples are shown for calculating irradiance values at different stages from top-of-atmosphere to at-sensor and at the land surface, using given pixel irradiance, gains, reflectance and atmospheric properties.

1. REMOTE SENSING
Thursday, 29 November 2017
J.Mwaura
RS APPLICATIONS & SEMINAR

2. Linear Contrast Stretching
0-255 in an 8-bit system
• 𝐷𝑁𝑠𝑡 = 255 × (𝐷𝑁 𝑖𝑛−𝐷𝑁 𝑚𝑖𝑛)
(𝐷𝑁 𝑚𝑎𝑥−𝐷𝑁 𝑚𝑖𝑛)
DN values range from 60-158
• 0-59 and 159-255 are not utilized

3. Histogram-equalized
Stretching
• Lets stretch values in the range [75-150]
– 𝐷𝑁𝑠𝑡 = 255 𝑗=0
𝑘
(
𝑛 𝑗
𝑁
)
• 𝐷𝑁𝑠𝑡 = 𝑒𝑛ℎ𝑎𝑛𝑐𝑒𝑑 𝐷𝑁 𝑣𝑎𝑙𝑢𝑒
• 𝑛𝑗 = 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑝𝑖𝑥𝑒𝑙𝑠 ℎ𝑎𝑣𝑖𝑛𝑔 𝐷𝑁 𝑣𝑎𝑙𝑢𝑒 𝑖𝑛 𝑡ℎ𝑒
𝑗 𝑡ℎ 𝑟𝑎𝑛𝑔𝑒, 𝑖𝑛 𝑡ℎ𝑒 𝑖𝑛𝑝𝑢𝑡 𝑖𝑚𝑎𝑔𝑒
• 𝑘 = 𝑚𝑎𝑥. 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝐷𝑁 𝑟𝑎𝑛𝑔𝑒 𝑖𝑛 𝑖𝑛𝑝𝑢𝑡 𝑖𝑚𝑎𝑔𝑒
• 𝑁 = 𝑻𝒐𝒕𝒂𝒍 𝒏𝒖𝒎𝒃𝒆𝒓 𝑜𝑓 𝑝𝑖𝑥𝑒𝑙𝑠 𝑖𝑛 𝑖𝑛𝑝𝑢𝑡 𝑖𝑚𝑎𝑔𝑒

4. Calculate Slope for
Band and Radiance
Metadata:
• LMAX_BAND6 = 15.303
• LMIN_BAND6 = 1.238
• 𝑆𝑙𝑜𝑝𝑒 =
(𝑙𝑚𝑎𝑥−𝑙𝑚𝑖𝑛)
𝑚𝑎𝑥
• 𝑆𝑙𝑜𝑝𝑒 𝐵6 =
15.303−1.238
255
• = 0.0551
• 𝑅𝑎𝑑𝑖𝑎𝑛𝑐𝑒 = 𝑠𝑙𝑜𝑝𝑒 × 𝐷𝑁 + 𝑙𝑚𝑖𝑛
• 𝑅 𝐵6 = 0.0551 × 𝐷𝑁 𝐵6 − 1.238
»e.g. DN = 69
• 0.0551 × 69 − 1.238 = 2.5639

5. Calculate Temperature
(at-sensor/TOA)
• Using Planck's Black Body Radiation Law
–𝑇 =
𝑘2
ln
𝑘1
𝑅
+1
,
𝑘 = 𝑐𝑎𝑙𝑖𝑏𝑟𝑎𝑡𝑖𝑜𝑛 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡,
𝑅 = 𝑟𝑎𝑑𝑖𝑎𝑛𝑐𝑒 𝑓𝑜𝑟 𝑏𝑎𝑛𝑑 = 2.5639
»𝑘1 = 666.09
»𝑘2 = 1282.71
• 𝑇𝐵6 =
1282.71
ln
666.09
2.5639
+1
= 195.54

6. • Calculating Surface Temperature
𝑇𝑠 =
𝑇𝐵
1 + λ ∗
𝑇𝐵
𝜌
∗ log 𝜀
– 𝑇𝐵 = 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒, 𝜀 = 𝑙𝑎𝑛𝑑 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑒𝑚𝑖𝑠𝑠𝑖𝑣𝑖𝑡𝑦
– λ = 𝑤𝑎𝑣𝑒𝑙𝑒𝑛𝑔𝑡ℎ, 𝜌 = ℎ 𝑝𝑙𝑎𝑛𝑐𝑘′ 𝑠 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 × 𝑐
𝜎 𝑏𝑜𝑙𝑡𝑧𝑚𝑎𝑛𝑛 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
• LSE (ε) can be extracted by using NDVI considering
three different cases
• Bare soil
• Vegetation
• Mixture of bare soil and vegetation

7. Land Surface
Emissity (LSE)
• For Band 6 of Landsat TM
• 𝜀 = 0.004 × 𝑝𝑣 + 0.986
• 𝑝𝑣 is the proportion of vegetation which is
given by
• 𝑝𝑣 = {
𝑁𝐷𝑉𝐼−𝑁𝐷𝑉𝐼 𝑚𝑖𝑛
(𝑁𝐷𝑉𝐼 𝑚𝑎𝑥−𝑁𝐷𝑉𝐼 𝑚𝑖𝑛)
}2

8. Land Surface
Temperature (LST)
• Parameters
»𝑇𝐵6 = 195.54
»𝜌 = 14380
»𝜀 = 1.2345
»𝜆 = 11.5 µm
– 𝑇𝑠 =
195.54
1+ 11.5×
195.54
14380
×log 1.2345
= 192.780
𝑐

9. Energy Interactions

10. Irradiance at-TOA?(1)
−6.70
𝐸 𝑅 = 𝐸 𝑇 × cos ∝
𝑇. 𝐶𝑎𝑛𝑐𝑒𝑟
𝐸𝑄𝑈𝐴𝑇𝑂𝑅
𝑇. 𝐶𝑎𝑝𝑟𝑖𝑐𝑜𝑛
𝑑 = 0.996
𝑆𝑎𝑡𝑒𝑙𝑙𝑖𝑡𝑒
𝐸 𝑇 =
𝐸𝑡
𝑑2
−23.30
?

11. Irradiance at-TOA?(2)
Sky
irradiance
(𝐼𝑠 = 12𝑢𝑛𝑖𝑡𝑠)
Path radiance
(𝑅 𝑝 = 2𝑢𝑛𝑖𝑡𝑠)
Pixel
irradiance
𝐼 𝑝 = 6𝑢𝑛𝑖𝑡𝑠
Sensor
𝐺𝑎𝑖𝑛 = 0.89,
𝑑𝑎𝑟𝑘 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 = −19
reflectance (𝜌 = 0.62)
Incident
irradiance (𝐸𝑖)
𝐓. 𝐎. 𝐀
Atmosphere
(t = 0.79)
Sky radiance
(𝑅 𝑠 = 14𝑢𝑛𝑖𝑡𝑠)
@𝐬𝐞𝐧𝐬𝐨𝐫
@𝐩𝐢𝐱𝐞𝐥

12. Irradiance at-TOA?(3)
1. 𝐸𝑡 =
258+19
0.89
= 311.24 𝑢𝑛𝑖𝑡𝑠
2. 𝐸𝑟 = 311.24 × cos(6.7) = 309.11 𝑢𝑛𝑖𝑡𝑠
 at Sensor Radiance
3. At pixel:
• 𝐼𝑠 = 0.79 × 12 = 9.48𝑢𝑛𝑖𝑡𝑠
• 𝐼 𝑝 = 0.79 × 6 = 4.74 𝑢𝑛𝑖𝑡𝑠

13. Irradiance at-TOA?(4)
4. Reflected at pixel:
• 𝐼𝑠 = 0.62 × 9.84 = 5.88𝑢𝑛𝑖𝑡𝑠
• 𝐼𝑠 = 0.62 × 4.74 = 2.94𝑢𝑛𝑖𝑡𝑠
5. At TOA (@sensor):
• 𝐼𝑠 = 0.79 × 5.88 = 4.65𝑢𝑛𝑖𝑡𝑠
• 𝐼 𝑝 = 0.79 × 2.94 = 2.32𝑢𝑛𝑖𝑡𝑠
• 𝑅 𝑝 = 0.79 × 2 = 1.58𝑢𝑛𝑖𝑡𝑠
• 𝑅 𝑠 = 0.79 × 14 = 11.06𝑢𝑛𝑖𝑡𝑠
6. Total = 19.61𝑢𝑛𝑖𝑡𝑠

14. Irradiance at-TOA?(5)
7. Reflected at pixel:
• 0.79 𝐸𝑟 + 19.61 = 309.11𝑢𝑛𝑖𝑡𝑠
• 𝐸𝑟 =
309.11−19.61
0.79
= 366.46𝑢𝑛𝑖𝑡𝑠
8. At TOA (@sun):
• 𝐸𝑖 =
366.46
0.62
= 591.06𝑢𝑛𝑖𝑡𝑠
• 𝐸𝑖@𝑡𝑜𝑎 =
591.06
0.79
= 748.17𝑢𝑛𝑖𝑡𝑠
• 𝐸𝑖@𝑡𝑜𝑎 =
748.17
0.9962 = 754.20𝑢𝑛𝑖𝑡𝑠

15. That’s it
Thanks…